SOLUTION: how do you solve arithmetic sequence problems.. find the sum of the terms of the arithmetic sequence. 1, 6, 11, 16, . . ., 116 n = 24

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Question 1094620: how do you solve arithmetic sequence problems..
find the sum of the terms of the arithmetic sequence.
1, 6, 11, 16, . . ., 116 n = 24
Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!

Because the terms in an arithmetic sequence are "equally spaced", the average of the first and last terms is equal to the average of all the terms. Since the sum of any set of number is the average of the numbers, multiplied by how many numbers there are, the sum of n terms of an arithmetic sequence is the number of terms, multiplied by the average of the first and last terms:

Sum = (# of terms) * (average of first and last)

For your example, there are 24 terms; the first is 1 and the last is 116. So
S+=+24%2A%281%2B116%29%2F2
S+=+12%2A117+=+1404

Note that the way I think of the sum of n terms of an arithmetic sequence is as described above: the number of terms, multiplied by the average of the first and last. If we call the first term a(1) and the n-th term a(n), then the formula I use is

S%28n%29+=+n%2A%28%28a%281%29%2Ba%28n%29%29%2F2%29

Many students prefer a slightly different way of thinking of the sum. They group the numbers in pairs, with the first pair being the first and last numbers, so that the sum in each pair is the same; then they multiply the number of pairs by that common sum. So the way they think of the formula is this:

S%28n%29+=+%28n%2F2%29%2A%28a%281%29%2Ba%28n%29%29

Of course the two formulas are equivalent -- it's just two different ways of viewing the process of finding the sum. Use whichever form of the formula "works" best for you.